Question 5 of 6: Restriction Enzymes and DNA Sequencing

Question 5: Restriction Enzymes and DNA Sequencing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) Whether the original DNA is linear or circular, with justification; (b) an estimated size for every fragment in all three digests; (c) a restriction map showing the relative positions of the A and B cleavage sites.

Approach. Use the relationship between the number of cut sites and the number of fragments produced: for LINEAR DNA, $n$ cut sites give $n+1$ fragments; for CIRCULAR DNA, $n$ cut sites give exactly $n$ fragments (cutting a circle just opens it up, it does not create an extra free end). Test both topologies against all three lanes simultaneously and keep only the interpretation that is consistent with every lane.

1. (a) Determine circular vs. linear. Enzyme B alone produces exactly ONE band. If the DNA were LINEAR, a single cut site would necessarily create TWO free ends and therefore TWO fragments (a lone cut essentially never lands exactly at a molecule's terminus) — so a truly linear molecule cut by an enzyme with one recognition site could not give a single band unless that enzyme has ZERO sites (i.e. does not cut at all, leaving the original uncut length as one band). But the A+B double digest clearly contains bands NOT present in the Enzyme-A-alone lane (an entirely new 3 kb and 1 kb pair replacing the 4 kb band), which proves Enzyme B DOES cut the molecule somewhere — ruling out the "B does not cut" linear explanation. The only topology consistent with "Enzyme B has at least one real cut site, yet produces just one band" is a CIRCULAR molecule: cutting a circular molecule at exactly one site does not fragment it at all, it simply linearizes the whole circle into a single linear piece equal to the FULL genome length — which is exactly why that band runs slower than (i.e. above/larger than) every sub-fragment seen in the other lanes, consistent with its position on the 10 kb tick, above the 8 kb marker. Conclusion: the digested DNA is circular.
2. (b) Fragment sizes and total genome length. For a circular molecule, $n$ cut sites give exactly $n$ fragments whose sizes sum to the total genome length. Enzyme A alone gives 2 fragments (6 kb and 4 kb, so 2 sites) summing to $$ 6 + 4 = \boxed{10\ \text{kb (total genome length)}} $$ Enzyme B alone gives 1 fragment (1 site) that, per Step 1, equals the full genome length — and its band sits on the 10 kb tick of the gel's size axis, exactly the 10 kb total from Enzyme A. The A+B double digest gives 3 bands (6, 3 and 1 kb), and $2\ (\text{from A}) + 1\ (\text{from B}) = 3$ total cut sites on the circle — consistent with 3 fragments. Their sum, $$ 6 + 3 + 1 = 10\ \text{kb} $$ matches the genome length from Enzyme A alone and from Enzyme B alone exactly, so all three lanes describe the same 10 kb circle.

   Question 5(b) — fragment sizes per digest

   Digest	Fragments (kb)	Sum (kb)
   A	6, 4	10
   B	10	10
   A + B	6, 3, 1	10

3. (c) Map the cleavage sites. Enzyme A's two sites (call them A1, A2) divide the circle into a 6 kb arc and a 4 kb arc. In the double digest, the 6 kb band reappears completely UNCHANGED, which means Enzyme B's single site does NOT fall inside that arc; instead it must fall inside A's other (4 kb) arc, splitting it into the two new fragments seen only in the A+B lane, 3 kb and 1 kb ($3+1=4$ kb, exactly the original A-fragment it replaces). B1 therefore lies 1 kb from one A site and 3 kb from the other. This gives a complete, self-consistent 3-site circular map, shown below. The data do not distinguish which of the two A sites the 1 kb piece sits against (A1 or A2) — resolving that would require an additional experiment (e.g. a partial digest, or double-digesting with B plus only one of the two A sites individually).

(d) Dideoxynucleoside triphosphates (ddNTPs) are the basis of Sanger dideoxy chain-termination sequencing. A ddNTP is chemically identical to a normal deoxynucleotide (dNTP) except that it lacks the 3'-hydroxyl group; DNA polymerase can still incorporate a ddNTP into a growing strand (base-paired correctly opposite the template), but because there is no 3'-OH for the next incoming nucleotide's phosphate to attack, chain elongation irreversibly stops at that position. In a sequencing reaction, a primer, template, all four normal dNTPs, DNA polymerase, and a small proportion of one (fluorescently or radioactively labelled) ddNTP are combined; because ddNTP incorporation competes stochastically with normal dNTP incorporation at every occurrence of that base along the template, the reaction produces a nested population of newly synthesized strands of every possible length, each one terminating exactly at a position where that particular base occurs. (Historically four separate reactions, one per base, were run in four lanes of a high-resolution polyacrylamide gel; modern capillary sequencing uses four different fluorescent dyes, one per ddNTP, in a single reaction.) Separating these fragments by size (shortest travels farthest) and reading off, in order of increasing fragment length, which base's ddNTP terminated each fragment directly reconstructs the DNA sequence, one base at a time, complementary to the template strand.